anonymous
  • anonymous
Consider the region under the curve y = 1/x, above the x axis, on the intervalbum 1 <= x < infinity. Calculate the volume of the solid obtained by revolving the region about the x axis.
Mathematics
katieb
  • katieb
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Astrophysics
  • Astrophysics
Astrophysics
  • Astrophysics
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Astrophysics
  • Astrophysics
\[V = \int\limits_{a}^{b} A(x) dx~~~~~~\text{where} ~ A(x) = \pi r ^2\]

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Astrophysics
  • Astrophysics
So I guess you can use the disk method, but it's been a while since I've done such problems, the radius for disk is f(x)
Astrophysics
  • Astrophysics
So \[V = \int\limits_{a}^{b} \pi f(x)^2 dx\]
Astrophysics
  • Astrophysics
Now just plug in the intervals and values and you're good to go!
anonymous
  • anonymous
Can you show me
Astrophysics
  • Astrophysics
\[V = \int\limits_{1}^{\infty} \pi \left( \frac{ 1 }{ x } \right)^2 dx\]
Astrophysics
  • Astrophysics
Go ahead and try to integrate now
anonymous
  • anonymous
the answer is 1.

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